Solve 30+13x-3x^2

$$30 + 13 x - 3 x ^ { 2 }$$

Factor

$$\left(6-x\right)\left(3x+5\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+13x+30$$

Factor the expression by grouping. First, the expression needs to be rewritten as -3x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=13 ab=-3\times 30=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -90.

$$-1,90 -2,45 -3,30 -5,18 -6,15 -9,10$$

Calculate the sum for each pair.

$$-1+90=89 -2+45=43 -3+30=27 -5+18=13 -6+15=9 -9+10=1$$

The solution is the pair that gives sum 13.

$$a=18 b=-5$$

Rewrite -3x^{2}+13x+30 as \left(-3x^{2}+18x\right)+\left(-5x+30\right).

$$\left(-3x^{2}+18x\right)+\left(-5x+30\right)$$

Factor out 3x in the first and 5 in the second group.

$$3x\left(-x+6\right)+5\left(-x+6\right)$$

Factor out common term -x+6 by using distributive property.

$$\left(-x+6\right)\left(3x+5\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$-3x^{2}+13x+30=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-13±\sqrt{13^{2}-4\left(-3\right)\times 30}}{2\left(-3\right)}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\left(-3\right)\times 30}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-13±\sqrt{169+12\times 30}}{2\left(-3\right)}$$

Multiply 12 times 30.

$$x=\frac{-13±\sqrt{169+360}}{2\left(-3\right)}$$

Add 169 to 360.

$$x=\frac{-13±\sqrt{529}}{2\left(-3\right)}$$

Take the square root of 529.

$$x=\frac{-13±23}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-13±23}{-6}$$

Now solve the equation x=\frac{-13±23}{-6} when ± is plus. Add -13 to 23.

$$x=\frac{10}{-6}$$

Reduce the fraction \frac{10}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{3}$$

Now solve the equation x=\frac{-13±23}{-6} when ± is minus. Subtract 23 from -13.

$$x=-\frac{36}{-6}$$

Divide -36 by -6.

$$x=6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{5}{3} for x_{1} and 6 for x_{2}.

$$-3x^{2}+13x+30=-3\left(x-\left(-\frac{5}{3}\right)\right)\left(x-6\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-3x^{2}+13x+30=-3\left(x+\frac{5}{3}\right)\left(x-6\right)$$

Add \frac{5}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-3x^{2}+13x+30=-3\times \frac{-3x-5}{-3}\left(x-6\right)$$

Cancel out 3, the greatest common factor in -3 and 3.

$$-3x^{2}+13x+30=\left(-3x-5\right)\left(x-6\right)$$

Evaluate

$$\left(6-x\right)\left(3x+5\right)$$